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The Orthogonal Colouring Game
Stephan Dominique Andres, Melissa Huggan, Fionn Mc Inerney, Richard
Nowakowski
To cite this version:
Stephan Dominique Andres, Melissa Huggan, Fionn Mc Inerney, Richard Nowakowski. The Orthogo-nal Colouring Game. Theoretical Computer Science, Elsevier, 2019, 795, pp.312-325. �hal-02017462v4�
The Orthogonal Colouring Game
Stephan Dominique Andresa, Melissa Hugganb,1, Fionn Mc Inerneyc,∗, Richard J. Nowakowskib,2
aFaculty of Mathematics and Computer Science, FernUniversit¨at in Hagen, IZ,
Universit¨atsstr. 1, 58084 Hagen, Germany
bDept. Math & Stats, Dalhousie University, Halifax, Canada
cUniversit´e Cˆote d’Azur, Inria, CNRS, I3S, 2004 route des Lucioles, 06902 Sophia
Antipolis, France
Abstract
We introduce the Orthogonal Colouring Game, in which two players alter-nately colour vertices (from a choice ofm ∈ N colours) of a pair of isomorphic graphs while respecting the properness and the orthogonality of the colour-ing. Each player aims to maximise her score, which is the number of coloured vertices in the copy of the graph she owns.
The main result of this paper is that the second player has a strategy to force a draw in this game for any m ∈ N for graphs that admit a strictly matched involution.
An involution σ of a graph G is strictly matched if its fixed point set induces a clique and any non-fixed pointv ∈ V (G) is connected with its image σ(v) by an edge. We give a structural characterisation of graphs admitting a strictly matched involution and bounds for the number of such graphs. Examples of such graphs are the graphs associated with Latin squares and sudoku squares.
Keywords: Orthogonal Colouring Game, orthogonal graph colouring,
∗Corresponding author
Email addresses: dominique.andres@fernuni-hagen.de(Stephan Dominique Andres), Melissa.Huggan@dal.ca (Melissa Huggan), fmcinern@gmail.com (Fionn Mc Inerney), r.nowakowski@dal.ca (Richard J. Nowakowski)
1This author is supported by the Natural Sciences and Engineering Research Council
of Canada and the Killam Trust.
2This author is supported by the Natural Sciences and Engineering Research Council
mutually orthogonal Latin squares, scoring game, games on graphs, strictly matched involution
2008 MSC: 05C57, 05C15, 05B15, 91A46, 91A43
1. Introduction
In this paper, we consider the following orthogonal colouring game, de-noted by M OCm(G). The board of the game consists of two initially
un-coloured disjoint isomorphic copies GA and GB of a given graph G. Two
players, Alice and Bob, with Alice beginning, alternately choose one of the two graphs GA or GB and colour an uncoloured vertex of this graph with
a colour from the set {1, . . . , m} such that adjacent vertices receive distinct colours (i.e., the partial colouring is proper ) and the orthogonality of the graphs is not violated. Orthogonality means that if v, w are two different vertices inG whose copies vA, wA(in GA) resp. vB, wB (in GB) are coloured,
then
(c(vA), c(vB)) 6= (c(wA), c(wB)), (1)
where c(x) denotes the colour of a vertex x. When no such move is possible any more, the game ends. Alice owns GA and Bob owns GB. The score of
a player is the number of coloured vertices in the board the player owns. If, at the end, the scores of both players are equal, there is a draw, otherwise, the player with the higher score wins.
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 GA GB GA GB GA GB GA GB GA GB GA GB GA GB
Figure 1: Alice’s winning strategy for the game M OC1(2K1)
The main result of this paper states that for a special class of graphs, graphs admitting a strictly matched involution, the second player, Bob, can achieve at least a draw.
This class of graphs includes many special cases where the game is to create combinatorial objects such as orthogonal Latin rectangles, double di-agonal Latin squares, Latin squares, and sudoku squares. However, there exist graphs in which optimal play from both players does not result in a draw. The smallest such example of a graph in which Alice wins is the graph 2K1 consisting of two isolated vertices with m = 1 colour, see Figure 1. An
example of a case where Bob wins is M OC2(C4). He wins as follows: when
Alice plays on her first move on a C4, Bob responds in the same C4 on the
non-adjacent vertex, colouring with the opposite colour if it is her C4 and
the same colour if it his C4. Bob will then win by 2 points. Note that Bob’s
optimal strategy is not to play just in his graph.
Moreover, in an accompanying paper, Andres et al. [2] proved that it is PSPACE-complete to determine the outcome of the orthogonal colouring game for any m ≥ 3 when an initial partial colouring is given.
Our paper is structured as follows. In Section 2, we motivate our research by the most prominent special case, the game played on the graph associated with Latin squares, and give references to related games and to some results on orthogonal graph colouring. In Section 3, we define graphs admitting a strictly matched involution and prove the main result of this paper. In Section 4, we study the graphs in which the game is a draw and prove for the most important special case, orthogonal Latin squares, that the game is a draw if m = 1. In Section 5, we provide a characterization of the graphs that admit a strictly matched involution which allows us to give an explicit construction for all such graphs and an upper bound for the number of these.
2. Motivation and Observations
The gameM OCm(G) emanates from the overlap of two lines of research:
combinatorial and scoring games (specifically, colouring games) and orthog-onality of Latin squares or, more generally, of colourings of graphs.
Combinatorial games have been vastly studied (e.g., the monographies of Albert et al. [1] or Berlekamp et al. [9], or, for a bibliography, the paper of Fraenkel [16]), particularly those where the board is a graph, e.g., by Beaudou et al. [6], Beeler [8], or Faridi et al. [15]. In general, the loser is the first player who cannot move—called the normal play convention. Note that the drawing strategy for the second player in the orthogonal colouring game is a winning strategy for the second player in the normal play convention. In an accompanying paper, Andres et al. [2] proved that it is PSPACE-complete
to determine the outcome of the game in the normal play convention when m ∈ N∗is the number of colours (even ifm is a fixed constant), and an initial
partial colouring is given.
The colouring game on graphs in the normal play convention, which was introduced as the Achievement game by Harary and Tuza [20] and called the Proper k-colouring game by Beaulieu et al. [7], is closely related to the orthogonal colouring game on graphs. In the Proper k-colouring game, two players take turns colouring the vertices of a graph, while main-taining that the colouring is proper. Beaulieau et al. [7] showed that this game is PSPACE-complete when k ∈ N∗ is the number of colours (even if k is a fixed constant), and an initial partial colouring is given. Fork = 1 colour, the Proper k-colouring game is the well-studied game Node Kayles. For specific classes of graphs, it is known which player wins the game Node Kayles, e.g., for paths and cycles a complete characterisation was given by Berlekamp et al. [9]. Harary and Tuza [20] characterised the winner in the Proper k-colouring game played with k = 2 colours on paths and cycles, and played with any number k of colours on the Petersen graph. Astonish-ingly, as far as we know, the Proper k-colouring game seems to not have been studied on other classes of graphs for k ≥ 2 colours.
More recently, there has been the development of a theory of scoring games by Larsson et al. [22, 23, 24] where the winner is the one with the greater score. Another type of scoring games, sometimes also called maker-breaker games, are based on the interplay of minimising versus maximising a score. Here, game-theoretic graph parameters are motivated by trying to get good approximations to graph parameters that are hard to calculate, e.g., chromatic number [17, 10, 14] and domination number [21]. Seo and Slater [26] give generic examples of how such parameters can be defined. Typically, two players choose vertices (or edges or other sub-objects) without violating a given property (e.g., independence). The score is the number of vertices chosen where one player wants to maximise the number and the other to minimise it. Note that, from the players’ point of view, it is difficult to determine the winner.
In particular, a game-theoretic version of the chromatic number, the game chromatic number, introduced by Bodlaender [10], and several of its possible variations have been extensively studied in the last three decades in more than 100 papers (see the partial surveys by Bartnicki et al. [5], Tuza and Zhu [27] or Dunn et al. [13] for some references). Upper bounds for the game chromatic number of many classes of graphs have been determined, e.g., for
forests by Faigle et al. [14], outerplanar graphs by Guan and Zhu [19], and planar graphs by Zhu [28]. However, the complexity of determining the game chromatic number of a graph in general is still an open problem.
Larsson et al. [25] extended the Maximiser/Minimiser approach, to use two graphs, G and H, usually, but not necessarily, isomorphic. One player, Left, is the maximiser on G but the minimiser on H and the other player, Right, has the reverse goals. The score is the number of pieces played on G minus the number played in H with Left winning if the score is positive and Right winning if the score is negative, and it is a draw if the score is 0.
Orthogonal colourings of graphs, i.e., proper colourings of two isomor-phic copiesGAandGB of a graph respecting the orthogonality condition (1),
have been studied as well (e.g., by Archdeacon et al. [3], Ballif [4], or Caro and Yuster [12]). Caro and Yuster [12] studied the parameters Oχ(G) and
Oχk(G) which are the minimum number of colours in any pair of
orthogo-nal colourings of G, respectively, required such that there exist k mutually orthogonal colourings of G. Specifically, the graph versions of combinatorial objects associated with orthogonality were studied by Ballif [4] such as Latin squares and Latin rectangles.
Orthogonal Latin squares are natural combinatorial objects where there are two ‘boards’ and these form the basis of a specific orthogonal colour-ing game played on Latin squares.
Recall (see Brualdi [11]) that ann × n square, partially filled with entries taken from {1, 2, . . . , n}, has the Latin property if each row and column does not contain any repeated entries. A fully filledn × n square is a Latin square if each entry is an integer between 1 andn (inclusive) and each row and each column contains all n integers, which implies that the square has the Latin property. For a (partially filled) n × n square, X, let cX(i, j) be the (i, j)
entry and ∅ if (i, j) is unfilled. Let A and B be (partially filled) n×n squares. Then A and B are orthogonal if in the list
((cA(i, j), cB(i, j)))1≤i≤n 1≤j≤n
every ordered pair of integers occurs at most once. If A and B are Latin squares, this means that every pair of integers from {1, . . . , n}2occurs exactly
once in the list.
Rules of the orthogonal Latin squares colouring game: There are two players, Alice and Bob, and two n × n squares labeled A and B. The squares are initially empty and Alice and Bob take turns filling one of the entries of either
A or B with an integer between 1 and m inclusive. After each move, both matrices must have the Latin property and the matrices must be orthogonal. It is known that a Latin square of order n can be regarded as a proper colouring of the cartesian product of Kn with itself. Thus, the concept of
or-thogonal Latin squares translates easily to graph colourings and the orthogo-nal colouring game played on Latin squares is equivalent toM OCm(KnKn).
See Figure 2 for an example of play.
3. Main Theorem
First, we fix some general notation. Forn ∈ N, let [n] := {1, . . . , n}.
We use standard notation from graph theory. The disjoint union of two graphsH and H0, denoted byH ∪ H0, is the graph (V ∪ V0, E ∪ E0) consisting
of an isomophic copy (V, E) of H and an isomorphic copy (V0, E0) ofH0 with
V ∩ V0 = ∅. The disjoint unionH ∪ H of two identical graphs is also denoted
by 2H.
Recall that, for a graph G = (V, E), an automorphism is a bijective mapping σ : V −→ V with the property that
∀v, w ∈ V : (vw ∈ E ⇐⇒ σ(v)σ(w) ∈ E). An involution of G is an automorphism σ of G with the property
∀v ∈ V : (σ ◦ σ)(v) = v. We define an involution of G to be strictly matched if
(SI 1) the set F ⊆ V of fixed points of σ (i.e., F = {v ∈ V | σ(v) = v}) induces a complete graph (i.e., for every v, w ∈ F with v 6= w we have vw ∈ E) and
(SI 2) for every v ∈ V \ F , we have the (matching) edge vσ(v) ∈ E.
If, for a graphG, there exists a strictly matched involution, we say that G admits a strictly matched involution. The following is the main result of this paper.
Theorem 1. Let G be a graph that admits a strictly matched involution and m ∈ N. Then, the second player has a strategy guaranteeing a draw in the game M OCm(G).
Board 1 Board 2 1 Alice’s move 1 Board 1 Board 2 1 1 Bob’s response 1 Board 1 Board 2 1 1 2 Alice’s move 2 Board 1 Board 2 1 1 2 2 Bob’s response 2 Board 1 Board 2 1 1 2 2 1 Alice’s move 3 Board 1 Board 2 1 1 2 2 1 1 OSP 1,1 Bob’s response 3 Board 1 Board 2 1 1 2 2 1 1 OSP 1,1 2 Alice’s move 4 Board 1 Board 2 1 1 2 2 1 1 OSP 1,1 2 2,2 2 Bob’s response 4 Board 1 Board 2 1 1 2 2 1 1 OSP 1,1 2 2,2 2 1 Alice’s move 5 Board 1 Board 2 1 1 2 2 1 1 OSP 1,1 2 2,2 2 1 1 Bob’s response 5 Board 1 Board 2 1 1 2 2 1 1 OSP 1,1 2 2,2 2 1 1 2 2,1 Alice’s move 6 Board 1 Board 2 1 1 2 2 1 1 OSP 1,1 2 2,2 2 1 1 2 2,1 2 1,2 Bob’s response 6
Figure 2: Bob’s strategy from the proof of Theorem 1 guarantees a draw in the orthogonal Latin squares colouring game: an example played on 3 × 3 squares with 2 colours.
In Figure 2, we illustrate Bob’s strategy given in the following proof of Theorem 1 on K3K3, where K3K3 is represented by a 3 × 3 board and
the involution is given by the mirror symmetry around the middle column of each board. Bob’s strategy on graphs in general just follows this idea using a strictly matched involution, as explained in the following.
Proof of Theorem 1. Let G1 and G2 be the two copies of G = (V, E). For
k ∈ {1, 2}, we denote by ck(v) the colour of the vertex v ∈ V in Gk. In case
the vertex v is uncoloured in Gk, we write ck(v) = ∅. To simplify notation
and differentiate between the colour of a vertex in a certain copy of G and an actual colour, we refer to the colours as symbols.
LetOSP be the set of orthogonal symbol pairs, i.e., the set of those pairs (s1, s2) of symbolss1, s2 ∈ [m], such that there exists a vertex v ∈ V with
c1(v) = s1 and c2(v) = s2.
Letσ be a strictly matched involution, which exists by precondition. For m = 0 or |V | = 0, the theorem is trivially true. Thus, assume m, |V | ≥ 1. The strategy of the second player, Bob, is to copy (in a certain sense) Alice’s moves in the other copy of the graph. Copying the symbols using the same positions would, in many cases, not be feasible because of orthogonality. Therefore, Bob couples the vertices of a graph with its image under σ of the other graph. Bob always plays the same symbol (=colour) as Alice just previously played.
Forc ∈ {c1, c2} we definec to be the other partial colouring from {c1, c2}
distinct from c.
Consider the case that Alice assignsc(v) := s for some c ∈ {c1, c2}, some
v ∈ V , and some symbol s ∈ [m]. Then, the copying strategy of Bob consists of assigning
c(σ(v)) := s.
We will prove that Bob will force a draw with this strategy.
We observe, as a key of our analysis, the following invariants which hold for every c ∈ {c1, c2}, every v ∈ V , and every s, s1, s2 ∈ [m] after each of
Bob’s moves:
1. Whenever c(v) = s, then c(σ(v)) = s. 2. Whenever c(v) = ∅, then c(σ(v)) = ∅.
3. Whenever (s1, s2) ∈OSP , then (s2, s1) ∈OSP .
We will prove by induction on the number of moves that after each move of Alice, Bob’s move assigningc(σ(v)) = s according to his strategy is possible, i.e.,
a) the vertex σ(v) is uncoloured in the colouring c; b) the move keeps the partial colourings being proper;
c) the move does not contradict the orthogonality of the colourings c1 of
G1 and c2 of G2;
and that after each move of Bob, the invariants hold again. At the beginning of the game, all invariants obviously hold.
Now consider a situation after a move of Alice, where she assignsc(v) = s. Therefore, before the move, vertex v was uncoloured, i.e., we had c(v) = ∅. By invariant 2, we had c(σ(v)) = ∅, thus, a) holds.
To prove b), assume to the contrary that the move of Bob would violate the properness of the partial colouringc, i.e., assume that there exists w ∈ V with w 6= σ(v) and wσ(v) ∈ E such that
c(w) = s = c(σ(v)).
As Alice played on the vertex v with the partial colouring c, the assignment c(w) = s must have been made before her move. Then, by invariant 1, we have c(σ(w)) = s. But, since wσ(v) ∈ E and σ is an involutive automor-phism, we have
σ(w)v = σ(w)σ(σ(v)) ∈ E,
which contradicts c(v) = s, since from w 6= σ(v) follows v 6= σ(w) by the properties of σ.
To prove c), we remark the following. In the case Alice has created a new element (s, x) ∈ OSP , then by invariant 4, (x, s) /∈ OSP before Alice’s move. Before proceeding with the proof, we will observe the following two key lemmas.
Lemma 2. For all x ∈ [m], at any time in the game, it is not possible for Alice to create a new element (x, x) ∈ OSP .
Proof. Assume Alice created a new element (x, x) ∈ OSP . Then, by the definition ofOSP , at some point in the game, a player has assigned c1(v1) =x
least one of these assignments was not performed in the last move w.r.t. the turn we consider. W.l.o.g. the assignment c1(v1) = x was performed before
the last move (the other case being symmetrical by interchanging the roles of G1 and G2). Then, by the invariants, the other player must have assigned
in the same pair of moves
c2(σ(v1)) =x. (2)
If v1 ∈ F (i.e., v1 is a fixed point of σ), then Bob created the new element
(x, x) ∈ OSP .
Otherwise, i.e., ifv1 ∈ V \F , as already mentioned above, by the definition
of OSP , at some point in the game a player has assigned
c2(v1) =x. (3)
But (2) and (3) contradict the facts that, by the definition (SI 2) of a strictly matched involution, there is a matching edgev1σ(v1) and, sincec2 is a proper
partial colouring, v1 and σ(v1) cannot be coloured the same colour.
Lemma 3. In case Bob must create a new element (x, x) ∈ OSP by his strategy, he is able to do so without violating orthogonality (i.e., (x, x) /∈ OSP before Alice’s move).
Proof. Assume (x, x) ∈ OSP before Alice’s move. Let v0 be the vertex with
c1(v0) = x = c2(v0). By invariant 1 and by orthogonality, v0 must be a fixed
point of σ (i.e., v0 ∈ F ). If Bob would have to create (x, x) ∈ OSP for the
second time, by his strategy, this would only be possible if Alice coloured a vertexv00∈ F , v00 6= v0, with colourx. But this is impossible, since, by (SI 1),
F induces a complete graph, thus, there is an edge v0v00 ∈ E, so that Alice
could not have colouredv00 with the same colour asv0. Thus, the assumption
is wrong. Therefore, Bob must create a new element (x, x) at most once. We continue with the proof of c). By Lemma 2, after Alice’s turn, we have
(s, x) 6= (x, s).
Therefore, also after Alice’s move, (x, s) /∈ OSP . Thus, the assignment c2(σ(v)) = s
of Bob is allowed (does not contradict orthogonality) and, since we have c1(σ(v)) = c2(v) = x, it will create a new element (x, s) ∈ OSP , satisfying
In case Alice has created a new element (x, s) ∈ OSP , the arguments are the same (just interchange the roles of c1 and c2).
In case Alice does not create a new element in OSP on her move, Bob will have a feasible move by invariant 3 and invariant 4, and by reasons of symmetry, will not create a new element in OSP unless Alice played some symbol s on a vertex in F , in which case, Bob creates the new element (s, s) ∈ OSP , which maintains invariant 3 and invariant 4. The latter move of player 2 is feasible because of Lemma 3. This proves c).
Now consider a situation after the move of Bob and assume a), b), and c) to be true before his move. We have to prove that the 4 invariants hold again.
Within the proof of c), we have shown that after Bob’s move, invariant 3 and invariant 4 hold again.
Invariants 1 and 2 follow from the definition of the assignment in Bob’s move and the induction hypothesis.
This concludes the inductive step.
We have shown that Bob’s strategy always allows a reaction to Alice’s move. Therefore, the game will end before a move of the first player. In such a situation, Bob’s copying strategy results in two partial colourings c1 and
c2 with exactly the same number of coloured vertices. Thus, the game ends
in a draw.
Corollary 4. The second player has a strategy to guarantee a draw in orthog-onal colouring games played on n1× n2 rectangles or n × n squares satisfying
the Latin property (and possibly the double diagonal condition or the sudoku condition).
To explain the notion used in Corollary 4: The double diagonal condition consists in demanding that the coloured entries of each of both diagonals in a square are pairwise different. The sudoku condition for an n × n square with n = k2 and
k ∈ N forces the coloured entries of each of the k2 disjoint
subsquares of size k × k to be pairwise different.
Proof of Corollary 4. For the graphs associated with such game boards, the assignment (i, j) 7−→ (i, n2 + 1 −j), which describes a vertical mirror
4. When the Game is a Draw
In this section, we look at graphs in general and also, at the special case of the graphs associated with Latin squares. We show that both players trivially have a strategy to draw ifm is large enough. For the game M OCm(KnKn),
we show that Alice has a strategy to draw if m = 1, thereby, showing that there exist graphs that admit a strictly matched involution where the optimal result for both players is a draw for some values of m.
First we note that, if m is large enough, both players have a strategy to force a draw. In the following lemma, for a graph G, the number ∆(G) is the maximum degree of a vertex in G and α(G) is the stability number (size of a maximum stable set) of G.
Lemma 5. For any graph G and all m ∈ N with m ≥ ∆(G) + α(G), both players have a strategy to draw in the M OCm(G) game.
Proof. We simply show that each of the players’ copies, G1 and G2, of the
graph G will be completely filled at the end of the game. To show this, we consider the worst possible case scenario for an uncoloured vertex v that needs to be coloured with some colours in some copy Gk ofG and show that
it is possible to colour vertex v with s in Gk. For k ∈ {1, 2}, consider an
uncoloured vertex v in some copy Gk of the graph G. Also, let k = 3 − k.
Thus, Gk is the other copy of the graph G that is not Gk.
In the case of the proper colouring property, the worst case is that every other vertex adjacent to v in Gk has been coloured with a distinct colour.
Then, ∆(G) colours are unavailable to be played on v in Gk by the proper
colouring property. See Figure 3 (a) for an example in the caseG = KnKn.
1 2 3 4 5 6 i j Gk
(a) Latin property and
i j 1 2 3 Gk i j a a a a Gk (b) orthogonality restrictions
Figure 3: In the case the graph G is the graph KnKn: the worst case in the proof of
By the orthogonality conditions, the worst case is that vertex v in Gk is coloured with some coloura and the forbidden pairs, (s1, s2) withb = sk and
a = sk, exist for some a ∈ [m] and α(G) − 1 values of b, where b ∈ [m]. Note that there cannot be more than α(G) − 1 colours unavailable for v by the orthogonality conditions. This is because the colour a that appears in v of Gk, may only appear at most α(G) times in Gk and hence, may only generate at most α(G) − 1 forbidden pairs with Gk, since v is not coloured
yet in Gk. See Figure 3 (b) for an example in the case G = KnKn.
To be precise, the worst case is that each of the α(G) − 1 unavailable colours by the orthogonality conditions for v in Gk, differ from each of the
∆(G) colours that are unavailable by the proper colouring property. Then, ∆(G) + α(G) − 1 colours are unavailable for v in Gk in the worst case.
Therefore, since m > ∆(G) + α(G) − 1, the vertex v of Gk may always be
coloured.
Corollary 6. For all m, n ∈ N with m ≥ 3n − 2, both players have a strategy to draw in the M OCm(KnKn) game.
Proof. By Lemma 5, the result follows from the facts that α(KnKn) = n
and ∆(KnKn) = 2n − 2.
Lemma 7. For all n ∈ N, both players have a strategy to guarantee a draw in the M OC1(KnKn) game.
Proof. By Theorem 1, Bob has a strategy to force a draw.
Now, we show that Alice has a strategy to force a draw. LetG be KnKn
and let Gk,k ∈ {1, 2}, be one copy of G and Gk be the other copy of G that
is not Gk. We identify the graphs G1 and G2 with their underlying square
boards. As we play with only m = 1 colour, it is easy to see that the only possible scores of both players aren and n − 1, regardless of strategy. This is due to the fact that the orthogonality condition can only block at most one vertex in a row or column from being coloured, as otherwise, it would violate the Latin property. Therefore, as long as more than one possible vertex exists for a row or column in Gk, then one of the vertices can be coloured. Thus,
the vertices of both Gk and Gk can be coloured until the point is reached
where two rows and two columns have no coloured vertices in them and at least one of these vertices can be coloured, guaranteeing a score of at least n − 1.
Alice, who owns copy G1 of G, colours a vertex in Bob’s copy, G2 of G,
vertices in G2, Alice colours a vertex inGk when Bob colours a vertex inGk.
Now, since Alice coloured a vertex in G2 initially, eventually it is Bob’s turn
and there aren − 3 coloured vertices in G1 and n − 2 coloured vertices in G2.
We show that Alice can force a draw from here. There are 3 cases based on the next move for Bob.
Case 1. Bob colours a vertex in G2 and there are no possible moves left
in G2.
In this case, Bob achieved a score ofn − 1 and so Alice can at least draw if not win.
Case 2. Bob colours a vertex in G2 and there is still a possible move left
in G2.
In this case, Alice colours the last colourable vertex in G2 and Bob
achieves a score of n. Bob is then forced to colour a vertex in G1 and now
it is Alice’s turn. There are two rows and columns in G1 with no coloured
vertices in them, and no more vertices may be coloured inG2. If none of the
4 colourable vertices remaining in G1 are already coloured in G2, then Alice
will clearly achieve a score of n. Otherwise, at most 2 of the 4 colourable vertices remaining in Alice’s board are already coloured in G2 and, by the
Latin property, they are not in the same row or column. Alice colours one of the 4 remaining colourable vertices in G1 that is in the same row or column
(but not the exact same position) as one of those at most 2 already coloured vertices in G2. Now it is not possible to stop Alice getting a score of n since
the last colourable vertex in G1 is not coloured in G2.
Case 3. Bob colours a vertex in G1.
BothG1 and G2 have two remaining rows and columns with no coloured
vertices in them. There are several cases of the possible situation. Let U1
(U2, respectively) be the set of the four possible remaining colourable vertices
inG1 (G2, respectively). Let [U1] and [U2] be the preimage ofU1 andU2 inG,
respectively. Note that at most two of the copies of the vertices in U1 may
already be coloured in G2 by the Latin property.
Subcase 3.1. 1 or 2 of the vertices in U1 have the property that their copies
are already coloured in G2.
If it is the case that two of the vertices in U1 have this property, then
Latin property would have been violated. Alice colours a vertex inU1 that is
in the same row or column (but not the exact same position) as one of these at most two already coloured vertices in G2. It is clearly not possible to stop
the last colourable vertex in G1 from being coloured eventually which results
in a score of n for Alice.
Subcase 3.2. None of the copies of the vertices in U1 have already been
coloured in G2.
• If [U1] ∩ [U2] = ∅, then clearly both players achieve a score of n.
• If |[U1] ∩ [U2]| ∈ {1, 2}, then clearly Alice has a strategy to get a score
of n by playing on a vertex in a position in [U1] ∩ [U2].
• The case |[U1] ∩ [U2]| = 3 is not possible.
• If [U1] = [U2], then Alice colours one of the vertices in U1. If Bob
colours a vertex in U1, then Alice achieves a score of n and so at least
draws the game if not wins. If Bob colours a vertex in U2 in the same
position or same column or row as the vertex Alice just coloured, then Alice can colour a vertex in U1 and achieves a score of n and again,
at least draws the game if not wins. Lastly, if Bob colours a vertex in U2 but not in the same position nor the same column or row as the
vertex Alice just coloured, then either Alice may still colour a vertex inU1 if there are no forbidden pairs due to orthogonality yet, in which
case Alice wins since Bob cannot colour a vertex in U2 on the next
turn by the orthogonality condition, or there is a forbidden pair due to orthogonality, in which case they draw with scores of n − 1 each since no vertices inU1norU2 can be coloured by the orthogonality condition.
5. Graphs that Admit a Strictly Matched Involution
We denote by MI the class of graphs that admit a strictly matched involution. See Figure 4 for a list of all graphs with at most 5 vertices that admit a strictly matched involution.
n = 0 n = 1 n = 2 n = 3 n = 4 n = 5
Figure 4: List of all graphs with ≤ 5 vertices that admit a strictly matched involution.
5.1. Characterising Graphs that Admit a Strictly Matched Involution
We first give an explicit characterization of all graphs G ∈ MI. We then use this characterization to give an explicit construction for any graph G ∈ MI. We note however, that in an accompanying paper, Andres et al. [2] proved that it is NP-complete to determine whether a graph admits a strictly matched involution.
Theorem 8. A graphG = (V, E) admits a strictly matched involution if and only if its vertex set V can be partitioned into a clique C and a set inducing a graph that has a perfect matching M such that:
1. for any two edges vw, xy ∈ M , the graph induced by v, w, x, y is iso-morphic to
(a) a 2K2 (2 disjoint copies of K2) or
(b) a C4 (there are two possibilities for this) or
(c) a K4;
2. for any edge vw ∈ M and any vertex z ∈ C, the graph induced by the vertices v, w, z is isomorphic to
(a) a K1∪ K2 or
Proof. First, we prove the forward implication of the theorem, that is, if a graph G = (V, E) admits a strictly matched involution, then the vertices V can be partitioned into a cliqueC and a matching M such that the properties (1.) and (2.) hold.
Thus, assumeG ∈ MI. Recall from the definition of a graph that admits a strictly matched involution, that (SI 1) and (SI 2) imply that the vertices V can be partitioned into a clique C and a matching M . Now, for any two edges vw, xy ∈ M , the graph induced by v, w, x, y ∈ V is isomorphic to either:
• a 2K2if no additional edges exist and note that this does not violate any
conditions in the definition of a graph that admits a strictly matched involution.
• or a C4 if vx and wy (vy and wx resp.) are edges in E or a K4 if
vx, wy, vy, wx ∈ E. Indeed, we prove that vx ∈ E if and only if wy ∈ E and vy ∈ E if and only if wx ∈ E, thereby proving that a C4 or a K4
are the only two possibilities if additional edges exist. We prove the first case as the other is analogous. Since G ∈ MI by assumption, and therefore, by (SI 2) and sinceσ is an involution, σ(v) = w and σ(x) = y, and since σ is an automorphism, vx ∈ E ⇔ σ(v)σ(x) = wy ∈ E. For any edge vw ∈ M and any vertex z ∈ C, the graph induced by v, w, z ∈ V is isomorphic to either:
• a K1 ∪ K2 if no additional edges exist and note that this does not
violate any conditions in the definition of a graph that admits a strictly matched involution.
• or a K3 ifvz, wz ∈ E. Indeed, we prove that vz ∈ E if and only if wz ∈
E, thereby proving that a K3 is the only possibility if additional edges
exist. The proof is analogous to the second case above and therefore, is omitted.
For the other implication, assume the vertices V can be partitioned into a clique C and a set inducing a graph that has a perfect matching M such that the properties (1.) and (2.) hold. We define a mapping σ as follows. For all vertices z ∈ C, let σ(z) = z and for all edges vw ∈ M , let σ(v) = w and σ(w) = v. We will prove that σ is a strictly matched involution.
Clearly, σ is involutive (i.e., σ(σ(v)) = v for every vertex v ∈ V ) and (SI 1) and (SI 2) are satisfied. Now all that remains to show is that σ is a graph homomorphism. That is, it remains to be proven that
vw ∈ E ⇐⇒ σ(v)σ(w) ∈ E. (4) First, the forward direction of (4) is proven. Let vw ∈ E. If vw ∈ M , then by our mapping, σ(v) = w and σ(w) = v and we are done. So, assume vw /∈ M . If v, w ∈ C, then σ(v) = v and σ(w) = w and we are done. So, w.l.o.g., assume that v /∈ C and let vx ∈ M . Then, σ(v) = x.
Ifw ∈ C, then σ(w) = w. Then, by property (2.), the graph induced by the verticesv, x, w is isomorphic to K3 (sincevw ∈ E and w ∈ C) and hence,
xw = σ(v)σ(w) ∈ E.
If w /∈ C, then let wz ∈ M . Then, σ(w) = z. Since M is a matching, z /∈ {v, x}. By property (1.), the graph induced by the vertices v, w, x, z is isomorphic to C4 or K4 (since vw ∈ E and vw /∈ M ) and in either case,
σ(v)σ(w) = xz ∈ E.
Using the forward direction and the fact thatσ is involutive, we immedi-ately get the backward direction of (4)
σ(v)σ(w) ∈ E =⇒ vw = σ(σ(v))σ(σ(w)) ∈ E. Thus, σ is strictly matched, i.e., G ∈ MI.
Theorem 8 immediately implies the following structural result.
Corollary 9. Any graph G on n vertices admitting a strictly matched in-volution has a partition of its vertex set into three (possibly empty) vertex subsets inducing a clique C of size n − 2k and two isomorphic graphs H and H0, each of size
k, for some k ∈ N with 0 ≤ k ≤n
2, respectively. Moreover,
• for any pair (v, v0) of corresponding vertices v ∈ V (H) and v0 ∈ V (H0)
and any vertex w ∈ C, either both vw and v0w exist or none of them; • for any pair (v, v0) of corresponding verticesv ∈ V (H) and v0 ∈ V (H0),
we have the existence of the matching edge vv0 ∈ E(G);
• for any two pairs (v, v0) and (w, w0) of corresponding vertices with
v, w ∈ V (H) and v0, w0 ∈ V (H0), either bothvw0 andv0w exist or none
of them.
Graph H of order k Clique Matching of size k Copy of H of size n−2k
Figure 5: The structure of graphs admitting a strictly matched involution
According to Corollary 9, we can generate every graph on n vertices admitting a strictly matched involution if we fix some integer k ≤ n
2 and
take two copies of an arbitrary graph onk vertices which are matched by an isomorphism and add possible edges according to the rules given implicitly in Theorem 8 and explicitly in Corollary 9. Note that this construction may create isomorphic and even identical graphs. However, it gives us an upper bound for the number of such graphs, as specified in the next theorem (Theorem 11).
5.2. Counting Graphs that Admit a Strictly Matched Involution
In the following, letg(n) be the number of isomorphism classes of graphs on n vertices. Let A(n) be the number of isomorphism classes of graphs admitting a strictly matched involution on n vertices.
We use the following well-known fact. Fact 10. For any n ∈ N,
2(n2)
n! ≤ g(n) ≤ 2(
n 2).
Theorem 11. For any n ∈ N, A(n) ≤ bn 2c X k=0 g(k)2(k2)2(n−2k)k.
Proof. To construct a graphG ∈ MI on n vertices, according to Corollary 9 (cf. Figure 5) we choose some nonnegative integer k with k ≤ n
2 and a
graph H on k vertices. By definition, for the latter, we have g(k) choices. We add an isomorphic copyH0ofH and a clique C on the remaining vertices.
For each pair of vertices ofH we have two choices for the edges between these vertices and their copies in the copy H0 of H: either there are only the two matching edges or all possible four edges exist. This gives us 2(k2) choices
for the edges between H and H0, since there are exactly k
2 such pairs.
For each pair (v, w) with v ∈ V (H) and w ∈ C, we have two choices: either vw ∈ E(G) or vw /∈ E(G). This gives us 2(n−2k)k choices for the
edges between H and the clique C. The edges between H0 and C are then
completely determined by these choices by Corollary 9.
Since all these choices are independent from each other, by summing over all k we get the claimed upper bound.
Corollary 12. For any n ∈ N, A(n) ≤ bn 2c X k=0 2k(n−1−k).
Proof. By Theorem 11 and the right hand inequality of Fact 10, we obtain
A(n) ≤ bn 2c X k=0 g(k)2(k2)2(n−2k)k ≤ bn 2c X k=0 2(k2)2( k 2)2(n−2k)k = bn 2c X k=0 2(n−1−k)k.
Corollary 13. For any n ∈ N,
A(n) ≤ 1 +jn 2 k · 2(n−12 ) 2 .
Proof. The exponent f (k) = (n − 1 − k)k in Corollary 12 is maximised for k = n−12 , therefore we conclude A(n) ≤ bn 2c X k=0 2(n−1−k)k = 1 + bn 2c X k=1 2(n−1−k)k ≤ 1 +jn 2 k 2(n−12 ) 2 .
Corollary 14. A(n) = Oc(n)pg(n) with log2(c(n)) = olog2pg(n)3
.
Proof. Using Corollary 13 and the left hand inequality of Fact 10 we obtain
A(n) ≤ 1 +jn 2 k 2(n−12 ) 2 = 1 +jn 2 k q 2(n2) p 2n−12 ≤ 1 + jn 2 k√ n! p 2n−12 pg(n). Then, c(n) := jn 2 k√ n! p 2n−12
is of the desired “moderately exponential ” form, since log√n! = O(n log n) (by Stirling’s formula) but logpg(n) = Ω (n3 2) (by Fact 10 and Stirling’s
formula).
There is also a trivial lower bound, given in the next theorem. Theorem 15. For any n ∈ N,
A(n) ≥ bn 2c X k=0 g(k)2(n−2k)k−jn 2 kn 2 .
Proof. We construct a set of pairwise non-isomorphic graphs that admit a strictly matched involution on n vertices. For every k = 0, 1, 2, 3, . . . ,n
2,
take any graphH on k vertices and an isomorphic copy H0 ofH, and connect
the vertices of H with their respective copy in H0 but with no other vertex
of H0, and form a clique C with the remaining vertices and choose any type
of connections between H and C and copy these connections between H0
and C. This gives us in total bn
2c
X
k=0
g(k)2(n−2k)k
graphs. We claim that all of these, except for at mostPb
n 2c k=1 n−2k 2 pairs, are non-isomorphic.
Since all of the graphs H are non-isomorphic, two of the constructed graphs can only be isomorphic if a part of the clique can be considered as
part of the matching. Since between theH and the H0 of such a graph there
are no edges apart from the matching edges, a part of the clique that is considered as part of the matching must consist of exactly two vertices v, w and these vertices must not be adjacent to any vertex in H and H0 (as we
forbid diagonal edges between H and H0 in our construction). For fixed k,
this gives us |C|2 = n−2k2 pairs of vertices, resp. graphs, that have already occurred. Summing up (and observing that the bound is exact for k = 0), we get at most bn 2c X k=1 n − 2k 2 ≤jn 2 kn 2
graphs for which an isomorphic graph was already constructed before. Corollary 16. For any n ∈ N,
A(n) ≥ bn 2c X k=0 1 k!2( n−1 2)k− 3 2k 2 −jn 2 kn 2 .
Proof. By Theorem 15 and the left hand inequality of Fact 10, we obtain
A(n) ≥ bn 2c X k=0 g(k)2(n−2k)k−jn 2 kn 2 ≥ bn 2c X k=0 1 k!2( k 2)2(n−2k)k− jn 2 kn 2 = bn 2c X k=0 1 k!2( n−1 2)k− 3 2k 2 −jn 2 kn 2 . Corollary 17. A(n) = Ωd(n)pg(n)3 with log2 1 d(n) =olog2pg(n)3 . Proof. The exponent f (k) = n −1
2 k − 3 2k 2 in Corollary 16 is maximised for k = n − 1 2 3 ,
therefore, for the integer valuei in the interval [(n − 2)/3, (n + 1)/3) we have f (i) ≥ f ((n − 2)/3).
Using this and the right hand inequality of Fact 10, we conclude from the lower bound in Corollary 16
A(n) ≥ bn 2c X k=0 1 k!2( n−1 2)k− 3 2k 2 −jn 2 kn 2 ≥ 1 i!2( n−1 2)i− 3 2i2 − jn 2 kn 2 ≥ 1 n!2( n−1 2)i− 3 2i2 − jn 2 kn 2 ≥ 1 n!2( n−1 2) n−2 3 − 3 2( n−2 3 ) 2 −jn 2 kn 2 = 1 n!2 1 3( n 2)− 1 3 − jn 2 kn 2 ≥ 3 pg(n) n!√3 2 − jn 2 kn 2 . Then, d(n) := 1 n!√3 2
is of the desired “moderately exponential ” form since log(n!√3
2) =O(n log n) (by Stirling’s formula) but
logpg(n) = Ω n3 2 (by Fact 10 and Stirling’s formula).
Improvements on the lower bounds seem possible since in our construction in Theorem 15 we used only one of the 2(k2) possible ways to connect H
with H0. We believe that the asymptotic behaviour of A(n) is nearer to the
upper bound than to the lower. Conjecture 18. Let m ∈ N. Then
A(n) = Θc(n)pg(n) with log2(c(n)) = olog2 mpg(n).
Conjecture 18 is justified by the observation that, in the construction of the graphs G ∈ MI in the proof of Theorem 11, relative to the total number of such graphs, not many pairs of graphs are isomorphic since most large graphsH have a trivial automorphism group (cf. Godsil and Royle [18, Chap. 2]).
6. Conclusion
We introduced a new scoring game on graphs called the orthogonal colour-ing game (M OCm(G)). We have shown for a large class of graphs, i.e., those
that admit a strictly matched involution, that Bob has a strategy to force a draw. We have also shown that Alice has a strategy to force a draw in a sub-class of this sub-class of graphs, when the game is played with certain numbersm of colours. For further work, it would be interesting to find another class of graphs in which one of the players wins or the game is a draw. Specifically, for graphs admitting a strictly matched involution, it would be interesting to know when there is a winning strategy for the second player.
Problem 19. Determine the outcome for the orthogonal colouring game for other classes of graphs.
Problem 20. For any m ∈ N, characterise the class of graphs that admit a strictly matched involution for which the game with m colours is a draw (second player win, respectively).
Acknowledgement
We would like to thank an anonymous referee for their suggestion to generalise Lemma 5.
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